Solution. With \(\Var (\varepsilon )=\sigma ^{2}\), \[\gamma (0) = \left (1+\theta ^{2}\right )\sigma ^{2},\qquad \gamma (1) = \theta \sigma ^{2},\qquad \gamma (\tau )=0 \ (\tau \geq 2),\] the last because \(X_t\) and \(X_{t+\tau }\) then share no shock. Dividing, \[\rho (1) = \frac {\theta }{1+\theta ^{2}},\qquad \rho (\tau )=0 \ (\tau \geq 2).\] Differentiating, \(\rho (1)\) is extremised at \(\theta =\pm 1\), giving \[\left |\rho (1)\right | \leq \tfrac 12 .\] So a sample first autocorrelation beyond \(0.5\) in magnitude rules out an MA\((1)\) outright, whatever \(\theta \) is chosen — a genuine identification constraint, not merely a guideline.

Note also that \(\theta \) and \(1/\theta \) give the same \(\rho (1)\): the autocorrelations do not determine the model, which is why invertibility is imposed to pick the root inside the unit circle.

Solution. Stationarity requires the roots of the characteristic equation \(1 - 0.5z - 0.3z^{2} = 0\) to lie outside the unit circle. Solving, \[z = \frac {0.5 \pm \sqrt {0.25 + 1.2}}{-0.6} = \frac {0.5 \pm 1.204}{-0.6},\] giving \(z \approx -2.84\) and \(z \approx 1.17\). Both exceed one in modulus, so the process is stationary.

The Yule–Walker equations give \[\rho (1) = \frac {\phi _1}{1-\phi _2} = \frac {0.5}{0.7} \approx 0.714 ,\] \[\rho (2) = \phi _1\rho (1) + \phi _2 = 0.5(0.714) + 0.3 \approx 0.657 .\] The slow decay is characteristic of an AR process: the ACF tails off rather than cutting, and it is the PACF that would cut off after lag two.

Solution. A trend is removed by first differencing, \(\nabla X_t = X_t - X_{t-1}\). Yearly seasonality in monthly data is removed by seasonal differencing at lag twelve, \(\nabla _{12}X_t = X_t - X_{t-12}\). Applying both, \[\nabla \nabla _{12}X_t = \left (X_t - X_{t-1}\right ) - \left (X_{t-12}-X_{t-13}\right ).\] Fitting an ARMA model to the result is a SARIMA\((p,1,q)\times (P,1,Q)_{12}\) model.

Two cautions. Differencing more than necessary inflates the variance and introduces artificial negative autocorrelation, so one difference of each kind is usually enough and the ACF should be inspected before applying a second. And a deterministic trend is better removed by regression than by differencing — the two are different models, and differencing a deterministic trend leaves a non-invertible moving-average term behind.

Solution. Uncorrelated residuals mean the model has captured the conditional mean correctly, so the ARMA orders need no revision. Correlated squared residuals mean the conditional variance is not constant — large errors follow large errors — which violates the homoscedasticity the ARMA model assumed.

The remedy is to add a variance model: fit an ARCH or GARCH term to the residuals, so the mean follows the ARMA structure and the variance follows the volatility structure. Ignoring it does not bias the point forecasts, but every interval built from the model will be wrong — too wide in quiet periods and far too narrow in turbulent ones, which is precisely when the interval matters.